A body of mass 8 kg performs SHM of amplitude 60 cm. The restoring force is 120N when the displacement is 60 cm. Its acceleration , PE and KE when displacement is 6 cm are

Options

A. `3 ms^(-2), 0.36 J,0`
B. ` 1.5 ms^(-2), 0.18 J ,0`
C. ` 1.5 ms^(-2) , 0.36 J , 0.18 J`
D. `15 ms^(-2) , 36J ,0`

Answer

To solve the problem step by step, we need to find the acceleration, potential energy (PE), and kinetic energy (KE) of a body performing simple harmonic motion (SHM) with the given parameters.

### Given Data:
- Mass of the body, \( m = 8 \, \text{kg} \)
- Amplitude, \( A = 60 \, \text{cm} = 0.6 \, \text{m} \)
- Restoring force, \( F = 120 \, \text{N} \) at displacement \( x = 60 \, \text{cm} = 0.6 \, \text{m} \)

### Step 1: Calculate the spring constant \( k \)
The restoring force in SHM is given by Hooke's Law:
\[ F = -kx \]
Taking magnitudes, we have:
\[ k = \frac{F}{x} \]
Substituting the values:
\[ k = \frac{120 \, \text{N}}{0.6 \, \text{m}} = 200 \, \text{N/m} \]

### Step 2: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is related to the spring constant \( k \) and mass \( m \) by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting the values:
\[ \omega = \sqrt{\frac{200 \, \text{N/m}}{8 \, \text{kg}}} = \sqrt{25} = 5 \, \text{rad/s} \]

### Step 3: Calculate the acceleration \( a \) at displacement \( x = 6 \, \text{cm} = 0.06 \, \text{m} \)
The acceleration in SHM is given by:
\[ a = -\omega^2 x \]
Taking magnitudes:
\[ a = \omega^2 x \]
Substituting the values:
\[ a = (5 \, \text{rad/s})^2 \times 0.06 \, \text{m} = 25 \times 0.06 = 1.5 \, \text{m/s}^2 \]

### Step 4: Calculate the potential energy (PE) at \( x = 6 \, \text{cm} \)
The potential energy in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
Substituting the values:
\[ PE = \frac{1}{2} \times 200 \, \text{N/m} \times (0.06 \, \text{m})^2 = \frac{1}{2} \times 200 \times 0.0036 = 0.36 \, \text{J} \]

### Step 5: Calculate the kinetic energy (KE) at \( x = 6 \, \text{cm} \)
The total mechanical energy (E) in SHM is given by:
\[ E = \frac{1}{2} k A^2 \]
Substituting the values:
\[ E = \frac{1}{2} \times 200 \, \text{N/m} \times (0.6 \, \text{m})^2 = \frac{1}{2} \times 200 \times 0.36 = 36 \, \text{J} \]
The kinetic energy can be found using:
\[ KE = E - PE \]
Substituting the values:
\[ KE = 36 \, \text{J} - 0.36 \, \text{J} = 35.64 \, \text{J} \]

### Final Results:
- Acceleration \( a = 1.5 \, \text{m/s}^2 \)
- Potential Energy \( PE = 0.36 \, \text{J} \)
- Kinetic Energy \( KE = 35.64 \, \text{J} \)